The range, R, of a projectile based on its angle of projection ∅ and velocity v can be directly calculated by:
R = (v²sin2∅)/g
Let the v of ball 1 be v₁
R₁ = (v₁²sin(2*30))/g
v₂ = 2v₁
R₂ = ((2v₁)²sin(2*30))/g
R₂ = 4[(v₁²sin(2*30))/g]
R₂ = 4R₁
So ratio of d2/d1 = 4
The electric field E of a charge is defined as E=F/Q where F is the Coulomb force and Q is the test charge.
E=(1/Q)*k*(q*Q)/r², where k=9*10^9 N*m²/C², q is the point charge, Q is the test charge and r is the distance between the charges.
So E=(k*q)/r²
When we input the numbers we get that electric field E of a point chage q is:
E=(9*10^9)*(5.4*10^-8)/0.2²=486/0.04=12150 N/C.
This is roughly E=12000 N/C =1.2*10^4 N/C
The correct answer is B.
Answer:
3.53×10⁶ N/c due west
Explanation:
From the question
E = F'/q........................ Equation 1
Where E = Electric Field, F = Net Force, q = Charge.
But,
F' = F₂-F₁...................... Equation 2
Substitute equation 2 into equation 1
E = (F₂-F₁)/q................ Equation 3
Given: F₁ = 3 N due east, F₂ = 15 N due west, q = 3.4×10⁻⁶ C
Substitute these values into equation 1
E = (15-3)/(3.4×10⁻⁶)
E = 12/(3.4×10⁻⁶)
E = 3.53×10⁶ N/c due west
